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3.cpp
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95 lines (78 loc) · 2.33 KB
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/*
3. Longest Substring Without Repeating Characters
https://leetcode.com/problems/longest-substring-without-repeating-characters/
Rank 3,684,591
Given a string s, find the length of the longest substring without repeating characters.
Example 1:
Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
Constraints:
0 <= s.length <= 5 * 10^4
s consists of English letters, digits, symbols and spaces.
Accepted 4.4M Submissions 13M Acceptance Rate 33.8%
*/
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <unordered_set>
#include <assert.h>
using namespace std;
typedef std::unordered_set<int > UoSet;
class Solution {
public:
int lengthOfLongestSubstring(string s) {
size_t maxlen = 0;
size_t start = 0;
UoSet set;
printf("------------ %s ---------------\n", s.c_str());
for (size_t i = 0; i < s.size(); i++) {
printf("%c\n", s[i]);
UoSet::iterator iter = set.find(s[i]);
printf("\t%lu. finding repeating: len=%lu\n", i, maxlen);
if (iter != set.end()) {
size_t len = i - start;
start = i;
maxlen = max(maxlen, len);
set.erase(iter);
printf("\t%lu. found: %c[%lu] set=%s\n", i, *iter, maxlen, set2string(set).c_str());
}
set.emplace(s[i]);
printf("\tset=%s\n", set2string(set).c_str());
}
printf("\t--> %lu \n", maxlen);
return maxlen;
}
string set2string(UoSet& set) {
string ret;
for (UoSet::iterator i = set.begin(); i != set.end(); i++) {
ret += *i;
}
return ret;
}
};
void test_case1() {
Solution s;
int ret = 0;
ret = s.lengthOfLongestSubstring("abcabcbb");
assert(ret == 3);
ret = s.lengthOfLongestSubstring("bbbbb");
assert(ret == 1);
ret = s.lengthOfLongestSubstring("pwwkew");
assert(ret == 3);
}
// c++ -Wall -g -std=c++11 3.cpp -o 3.out
int main() {
test_case1();
return 0;
}